
\prob{008E}{根式方程IV}

若$x\sqrt{2 - y^2} + y\sqrt{2 - x^2} = 2$，求$x^2 + y^2$。
\problabels{yellow/代数, green/方程相关问题}

\ans{$x^2 + y^2 = 2$}

\subsection{配方}

两边乘以$-2$，移项得
\[ 4 - 2x\sqrt{2 - y^2} - 2y\sqrt{2 - x^2} = 0 \]
于是有
\begin{align*}
  x^2 - 2x\sqrt{2 - y^2} + \left(2 - y^2\right) & \\
  + y^2 - 2y\sqrt{2 - x^2} + \left(2 - x^2\right) &= 0 \\
  \left(x - \sqrt{2 - y^2}\right)^2 + \left(y - \sqrt{2 - x^2}\right)^2 &= 0 \\
\end{align*}
于是有$x = \sqrt{2 - y^2} \Rightarrow x^2 = 2 - y^2$，代入得
\[ x^2 + y^2 = \left(2 - y^2\right) + y^2 = 2 \]
故$x^2 + y^2 = 2$。
